Question

Find the value of k for which the following pair of linear equations has infinitely many solutions. 
2x + 3y = 7, (k +1) x+ (2k -1) y = 4k + 1

Answer 1

We have,

`2x + 3y = 7 ⇒ 2x + 3y - 7 = 0`
`(k + 1) x + (2k - 1)y = 4k + 1 ⇒ (k + 1)x (2k -1)y - (4k + 1) = 0`

For infinitely many solutions

`a_1/a_2 = b_1/b_2 = c_1/c_2`

⇒ `(2)/(k+1) = (3)/((2k -1)) = (-7)/-(4k +1)`

⇒ `(2)/(k+1) = (3)/(2k -1)`

⇒ `2(2k + 1) = 3 (k+1)`

⇒`4k - 2 = 3k + 3`

⇒`4k - 3k = 3 +2`

`k = 5`

or

⇒ `(2)/(k+1) = (-7)/-(4k +1)`

⇒ `2(4k + 1) = 7 (k+1)`

⇒ `8k + 2 = 7k + 2`

⇒`8k - 7k = 7- 2`

`k = 5`

Hence, the value of k is 5 for which given equations have infinitely many solutions.